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Juliette [100K]
2 years ago
7

The relation R is shown below as a list of ordered pairs. R = { (1, 4), (1, 3), (-1, 3), (2, 15) } Which ordered pairs prevent t

his relation from being a function? (1, 4) and (1, 3), because they have the same x-value (1, 3) and (–1, 3), because they have the same y-value
Mathematics
2 answers:
s344n2d4d5 [400]2 years ago
8 0

Answer:

(1,4) and (1,3)

Step-by-step explanation:

A function may not have two y-values assigned to the same x-value, it may have two x-values assigned to the same y-value

Therefore the order pair (1,4) and (1,3) prevents this relation from being a function.

Each x has only one y-value.

Sindrei [870]2 years ago
8 0

Answer: Points (ordered pairs) in a function are allowed to have the same y-value. If some do, they prevent the relation from being one-to-one, but do not prevent the relation from being a function.


(1,4) and (1,3) prevent the relation from being a function.


After you remove one of them, pairs (1,3) and (-1,3) prevent the function from being one to one, which means it has no inverse, but it is a function.


Step-by-step explanation:


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EXPLANATION

Method 1: Rewrite as product of
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The expression given to us is,

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{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


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{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


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{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


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